$g(x)=\dfrac{1}{1+x^2}$ We know that $g(x)=1-x^2+x^4-x^6+...$ for $x\in(-1,1)$. Using this fact, find the power series for $f(x)=-\dfrac{2x}{(1+x^2)^2}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-2x+4x^3-6x^5+...$ (Choice B) B $1-2x+4x^3-6x^5+...$ (Choice C) C $-2x-4x^3-6x^5+...$ (Choice D) D $2x+4x^3+6x^5+...$ (Choice E) E $1+2x+4x^3+6x^5+...$
Solution: First, notice that the derivative of $ ~g(x)=\frac{1}{1+x^2}~~$ is $~~-\frac{2x}{{{(1+x^2)}^{2}}}\,$. This is precisely the function $~f(x)~$ that we are looking for. From our knowledge of geometric series, we know how to expand $~g(x)=\dfrac{1}{1+x^2}~$ as a series. $\frac{1}{1+x^2}=1-x^2+{{x}^{4}}-{{x}^{6}}+{{x}^{8}}+...$ Taking the derivative of both sides gives us the solution to the problem. $-\frac{2x}{{{(1+x^2)}^{2}}}=-2x+4{{x}^{3}}-6{{x}^{5}}+...$